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Numerade Educator

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Problem 26 Medium Difficulty

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.

$ g(t) = \dfrac{1}{\sqrt{t}} $

Answer

$g^{\prime}(t)=\frac{-1}{2 t \sqrt{t}}$
Domain of $g(t)$ and $g^{\prime}(t)$ is $x \in(0, \infty)$

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Video Transcript

given GFT which is one over square root of tea. We need to find G prime of T. Using only derivative definition. Now by definition of derivatives we have G prime of T. This is equal to The limit as h approaches zero of G of T plus h minus guilty this all over each. And so from here we have limits As it approaches zero of 1 over the square root of T plus H -1 over the square of T. This times The reciprocal of a church is one over H. And then from here we want to combine the two fractions that's going to be limit as h approaches zero of we have an LCD of square it of T plus each time squared of teeth. And the numerator should be squared of t minus squared of T passage this times one over each. And then from here we want to rationalize the numerator. We multiply the numerator and denominator by the conjugate that is squared of T plus squared of T plus H. This all over the same expression squared of T plus square it of T plus H. And so simplifying this. We have The limit as h approaches zero of we have squared of tea and then squared minus the square of the squared of sleepless age. This all over we have squares a sleepless age times a squared of tee, times H times the congregate squared of T plus squared of T plus H. I'm simplifying this. We have they met H approaches zero of the numerator becomes t minus T plus H this all over. Yes scared of T plus H. Time squared of T. Times H times squared of T plus squared of T passage. Simplifying further we have in here this becomes minus t minus H. And so this simplifies to negative H. Because the tea will cancel. And so we have limits S. H approaches zero of negative H. Over we have squares of C. Plus age times the square root of T. Times age times squared of T plus squared of T passage. And then from here we can cancel the H. Reduce it to one. And so we have limit as H approaches zero of negative of one over square it of T plus H. Times squared of T. Times the square of T plus the square of T plus H. and then evaluating at age we get negative one over Square of TP0 that's squared of T. Times squared of T. Times we have squared of T. Plus scared of T. Plus zero. That's just squared of T. And so we have negative one over scared of tea time scared of T. That's just T. And then skirt of T. Plus scared of T. That's two square it of T. And so we have negative one over to T squared of T. And so this is the derivative of the function for the domain of G. F T. And G. Prime of T. F G F T Is one over the square root of T. Then T. Is restricted. such that T must be greater than zero only. It cannot equal zero because T. Is or the squared of T is in the denominator where it cannot be zero there, so it's only T greater than zero, or an interval notation. This is The interval from 0 to positive infinity, Where zero is not included. And if G prime of T is equal to negative one over to T squared of T, then it will have the same domain as the function because squared S. T. This is restricted. Such that team must be greater than zero because script of these in the denominator and the inside the squares must be positive. And so the domain must be The same 0 to positive infinity. zero is not included.